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Nat2105 [25]
3 years ago
7

When does a skydiver achieve terminal velocity?

Physics
1 answer:
LekaFEV [45]3 years ago
7 0
Based on wind resistance
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A satellite in outer space is moving at a constant velocity of 20.5 m/s in the +y direction when one of its on board thruster tu
Katyanochek1 [597]

Answer:

a)  V_f = 25.514 m/s

b)  Q =53.46 degrees CCW from + x-axis

Explanation:

Given:

- Initial speed V_i = 20.5 j m/s

- Acceleration a = 0.31 i m/s^2

- Time duration for acceleration t = 49.0 s

Find:

(a) What is the magnitude of the satellite's velocity when the thruster turns off?

(b) What is the direction of the satellite's velocity when the thruster turns off? Give your answer as an angle measured counterclockwise from the +x-axis.

Solution:

- We can apply the kinematic equation of motion for our problem assuming a constant acceleration as given:

                                   V_f = V_i + a*t

                                   V_f = 20.5 j + 0.31 i *49

                                   V_f = 20.5 j + 15.19 i

- The magnitude of the velocity vector is given by:

                                   V_f = sqrt ( 20.5^2 + 15.19^2)

                                   V_f = sqrt(650.9861)

                                  V_f = 25.514 m/s

- The direction of the velocity vector can be computed by using x and y components of velocity found above:

                                 tan(Q) = (V_y / V_x)

                                 Q = arctan (20.5 / 15.19)

                                 Q =53.46 degrees

- The velocity vector is at angle @ 53.46 degrees CCW from the positive x-axis.

4 0
3 years ago
You are standing a distance of 17.0 meters from the center of a merry-go-round. The merry-go-round takes 9.50 seconds to go comp
GenaCL600 [577]

Answer:

(a)11.24 m/s

(b)7.44 m/s

(c)409 N

(d)539.55\mu

(e) 0

Explanation:

The period for 1 circle 2\pi of the merry go around is 9.5s. It means the angular speed is:

\omega = \theta / t = 2\pi / 9.5 \approx 0.661 rad/s

(a)The speed is

v = \omega * R = 0.661 * 17 = 11.24 m/s

(b) Centripetal acceleration:

a = \frac{v^2}{R} = \frac{11.24^2}{17} = 7.44 m/s^2

(c) Magnitude of the force that keeps you go around at this acceleration

F = ma = 55 * 7.44 = 409 N

(d) let the coefficient of friction by \mu. The frictional force shall be this coefficient multiplied by normal force reverting gravity of the man

F_f = mg\mu = 55*9.81\mu = 539.55\mu

3 0
3 years ago
A photograph is taken by letting light fall on a light-sensitive medium, which then records the image onto that medium. True or
Jet001 [13]

Answer;

The above statement is true.

-A photograph is taken by letting light fall on a light-sensitive medium, which then records the image onto that medium.

Explanation;

-A photograph is created when light is allowed to fall on a light-sensitive medium. The pattern of light creates an image that is recorded by the photographic device. How light or dark a photograph is depends on how much light was allowed to fall on the light-sensitive medium.

-A camera is a light-tight box that contains a light-sensitive material or device and a way of letting in a desired amount of light at particular times to create an image on the light-sensitive material.

6 0
3 years ago
Imagine two billiard balls on a pool table. Ball A has a mass of 7 kilograms and ball B has a mass of 2 kilograms. The initial v
olga_2 [115]
A) the final velocity = 66/9 m/s.
b) The total momentum before and after collision is the same because energy is destroyed or made.
Thanks brainly. <span />
7 0
4 years ago
How do electrons flow through a wire
aniked [119]

Every atom has electrons. When you add new electrons to the wire, they will be passed on to an atom. The electrons keep passing from atom to atom until it reaches the light source, basically. It's kinda like that one song "100 jugs of milk" or whatever it's called. Each atom passes the atom next to it an electron.

7 0
3 years ago
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